Q67P - University Physics with Modern Physics

Question

The sound source of a ship’s sonar system operates at a frequency of 18.0 kHz. The speed of sound in water (assumed to be at a uniform 20°C) is 1482 m>s. (a) What is the wavelength of the waves emitted by the source? (b) What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling directly toward the ship at 4.95 m/s? The ship is at rest in the water.

Step-by-Step Solution

Answer

A) The wavelength of wave is 0.0823 m b) The difference in frequency of radiated wave is 120.6 Hz

STEP 1 Concept of the wavelength of the wave

The wavelength of wave is given as $\lambda =vf$were, $\lambda$ is the wavelength of wave, v is the speed of sound and f is the frequency of source.                                                                                  

 The difference in frequency of radiated wave is $\mathrm{\Delta f}=2\mathrm{vwv}-\mathrm{vwf}$ $∆f$is the difference in frequency of radiated wave and vw is the velocity of whale

STEP 2 Calculate the wavelength of wave

The frequency of sound source is f=18.O kHz, The speed of sound in water is v=1482 m/s. the wavelength of wave is given as $\lambda =vf$. Substitute the known value in the formula to calculate the wavelength of wave.

$$\begin{gathered} \lambda =1482\times 1800 \\ =00823\mathrm{m} \end{gathered}$$

Therefore, the wavelength of wave is 0.0823 m

STEP 3 Calculate the difference in frequency of radiated wave

The difference in frequency of radiated wave is $\mathrm{\Delta f}=2\mathrm{vwv}-\mathrm{vwf}$  The velocity of whale is v w=4.95 m/s. Substitute the known value in the formula to calculate the difference in frequency of radiated wave

$$\begin{gathered} ∆\mathrm{f}=2495\mathrm{x}1482-4.95\mathrm{x}18.0 \\ =120.6\mathrm{Hz} \end{gathered}$$

Thus, the difference in frequency of radiated wave is 120.6 Hz